求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期
来源:学生作业帮 编辑:大师作文网作业帮 分类:综合作业 时间:2024/09/20 05:53:09
求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期
y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]
=[sin2x+sin2xcosπ/3+cos2xsinπ/3]/[cos2x +cos2xcosπ/3-sin2xsinπ/3]
=[sin2x+1/2*sin2x+√3/2*cos2x]/[cos2x +1/2*cos2x-√3/2*sin2x]
=[3/2*sin2x+√3/2*cos2x]/[3/2*cos2x-√3/2*sin2x]
=√3(√3/2*sin2x+1/2*cos2x)/[√3(√3/2*cos2x-1/2*cos2x)]
=√3(sin2xcosπ/6+cos2xsinπ/6)/[√3(cos2xcosπ/6-sin2xsinπ/6)]
=(sin2xcosπ/6+cos2xsinπ/6)/(cos2xcosπ/6-sin2xsinπ/6)
=sin(2x+π/6)/cos(2x+π/6)
=tan(2x+π/6)
最小正周期T=π/w=π/2
=[sin2x+sin2xcosπ/3+cos2xsinπ/3]/[cos2x +cos2xcosπ/3-sin2xsinπ/3]
=[sin2x+1/2*sin2x+√3/2*cos2x]/[cos2x +1/2*cos2x-√3/2*sin2x]
=[3/2*sin2x+√3/2*cos2x]/[3/2*cos2x-√3/2*sin2x]
=√3(√3/2*sin2x+1/2*cos2x)/[√3(√3/2*cos2x-1/2*cos2x)]
=√3(sin2xcosπ/6+cos2xsinπ/6)/[√3(cos2xcosπ/6-sin2xsinπ/6)]
=(sin2xcosπ/6+cos2xsinπ/6)/(cos2xcosπ/6-sin2xsinπ/6)
=sin(2x+π/6)/cos(2x+π/6)
=tan(2x+π/6)
最小正周期T=π/w=π/2
求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期
求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期
y=sin(π/3-2x)+cos2x求函数最小正周期
1.y=cos^4x+sin^4x 求周期 2.y=(sin2x+sin(2x+π/3))/( cos2x+cos(2x
求函数Y=sin^4x +cos^4x +sin^2x cos^2x 除以2-sin2x的最小正周期,最大值最小值
函数y=sin(π/6-2x)+cos2x的最小正周期为
函数y=3sin^2x+cos2x的最小正周期为?
函数y=sin2x+2√3sin^2x的最小正周期T为
函数y=sin(派/3-2x)+sin2x的最小正周期是?
已知函数f(x)=cos(x-π|3)-sin(π|2-x)求函数的最小正周期
函数y=根号3cos^2x+1/2sin2x的最小正周期怎么求
已知函数f(x)=sin(2x+π/6)+sin2x-π/6+2cos²x求函数的最小值及最小正周期